Let $f:R\to R$ be a continuous function such that $f\left(x\right)+f\left(x+1\right)=2$ for all $x\in R$. If $I_1={\int }_0^{8}f\left(x\right)dx$ and $I_2={\int }_{-1}^{3}f\left(x\right)dx$, then the value of $I_1+2I_2$ is equal to ________.

$f\left(x\right)+f\left(x+1\right)=2$

Replace $x \text{by} x+1$

$f\left(x+1\right)+f\left(x+2\right)=2$

By subtracting the equations, we get

$f\left(x\right)=f\left(x+2\right)$

$\Rightarrow f\left(x\right)$ is periodic with period $=2$

$I_1={\int }_0^{8}f\left(x\right)dx=4{\int }_0^{2}f\left(x\right)dx$

$=4{\int }_0^1\left(f\left(x\right)+f\left(1+x\right)\right)dx=8$

Similarly, $I_2=2\times 2=4$

Hence, $I_1+2I_2=16$